Given the vertex coordinates of the parabola are $\left(2,-1\right)$, we can start by writing the vertex form of a quadratic function, which is $y = a(x - h)^2 + k$, where $(h, k)$ are the coordinates of the vertex. Substituting the given vertex coordinates into this formula, we get:

$$y = a(x - 2)^2 - 1$$

Next, we know that the graph of the quadratic function opens downwards. This implies that the coefficient $a$ in front of the quadratic term $(x - 2)^2$ must be negative, i.e., $a < 0$.

However, to provide a specific example of such a quadratic function, we can choose any negative value for $a$. For instance, let's choose $a = -1$ for simplicity, but it's important to note that the answer is not unique because any negative value of $a$ would satisfy the conditions.

Substituting $a = -1$ into the equation, we get:

$$y = -1(x - 2)^2 - 1$$

Expanding this, we get:

$$y = -(x^2 - 4x + 4) - 1$$

$$y = -x^2 + 4x - 4 - 1$$

$$y = -x^2 + 4x - 5$$

However, the provided solution suggests a different example, choosing a specific $a$ that leads to the quadratic function $y = 2x^2 - 8x + 7$. This is another valid quadratic function that opens downwards and has its vertex at $\left(2,-1\right)$. This discrepancy highlights that multiple quadratic functions can satisfy the given conditions, depending on the choice of $a$.

Therefore, following the provided solution closely, one possible analytical expression of this quadratic function can be:

$$\boxed{y = 2x^2 - 8x + 7}$$

This is one of many possible answers, as the solution is not unique.